Question

A metallurgist has an alloy with 5​% titanium and an alloy with 25​% titanium. He needs 100 grams of an alloy with 13​% titanium. How much of each alloy should be mixed to attain the 100 grams of alloy with 13% ​titanium

Answer 1

So... the alloy with 5% titanium is ...some  alloy :), so is a mixed metal and among its mix it has 5% of titanium, it has other metals, but only 5% is titanium.

that means that if the amount of it is say "x", then the amount of titanium in it will be 5% or (5/100) * x, or 0.05x.

now, the 25% titanium, is the same issue, if say a quantity of "y" will only have 25% of titanium, regardless of other metals, so in "y" there is 25% or (25/100) * y, or 0.25y.

so, we need a mixture that yields only a 13% alloy, or 13/100, and will be 100 grams, so the amount of titanium in it will then be (13/100) * 100 or 13.

$\bf \begin{array}{lccclll} &\stackrel{grams}{amount}&\stackrel{\%}{quantity}&\stackrel{concentrated}{quantity}\\ \textit{5\% alloy}&x&0.05&0.05x\\ \textit{25\% alloy}&y&0.25&0.25y\\ ------&------&------&------\\ mixture&100&0.13&13 \end{array} \\\\\\ \begin{cases} x+y=100\implies \boxed{y}=100-x\\ 0.05x+0.25y=13\\ ----------\\ 0.05x+0.25\left( \boxed{100-x} \right)=13 \end{cases} \\\\\\ 0.05x-0.25x+25=13\implies -0.20x=-12\implies x=\cfrac{-12}{=0.20} \\\\\\ x=\stackrel{grams}{60}$

how much of the 25% alloy will be needed?  well, y = 100 - x.